Let $E$ be a ($\Bbb{R}$ or $\Bbb{C}$) finite dimension vector space.
Prove that an endomorphism $f$ of $E$ is diagonalizable if and only if there exists numbers $(a_1,..a_n$), endomorphisms $(u_1,..,u_n)$ such that $\forall k \in 0..n:$ $$f^k=\sum_{i=1}^{n}a_i^ku_i$$
If $f$ is diagonalizable then the result is evident as we can take $n=dim(E)$ and $(a_1..a_n)$ the eigenvalues of $f$.
But I'm having real trouble with the other way around. What approach should I take? try to prove that the $a_i$ are necessarily the eigenvalues? We don't even know if the $(a_i)$ are distinct. (If they are, I think we can write $(u_i)$ as a polynomial of $f$ using a Vandermonde matrix so they commute but then what?)
Any help would be welcome
Edit: I think I got it.
for any polynomial $P(X)=\sum_{k=0}^n p_kX^k$:
$P(f) = \sum_{k=0}^n p_ku^k = \sum_{k=0}^np_k\sum_{i=1}^na_i^ku_i = \sum_{i=1}^n(\sum_{k=0}^np_ka_i^k)u_i = \sum_{i=1}^nP(a_i)u_i$
So taking the polynomial $P(X)=\prod_{i=1}^n (X-a_i)$ we have $P(f) = 0$ and $P$ has simple zeros.